What is the volume of diborane in this reation?

Question

Diborane, B2H6, is a highly explosive compound formed by the reaction:
3NaBH4 + 4BF3 —> 2B2H6 + 3NaBF4 (already balanced)
What volume of diborane at 22°C and 738 torr is formed from 26.0 g NaBH4 (sodium borohydride)?

Answer 1

First determine the number of moles of diborane, then use the ideal gas equation to determine the volume:

Sodium borohydride weighs:

[23 + 10.8 + (4 x 1.01)]g/mol = 37.84g/mol

26.0g/37.84g/mol = 0.687 mole of NaBH4

for each 3 moles of sodium borohydride you get 2 moles of diborane, so;

0.687mole NaBH4 x (2/3) = 0.458mole B2H6

PV = nRT solving for V: V = nRT/P

Get things in proper units:

22C = 295K; 738torr = 0.971atm

Plug in the values to get V:

[0.458mole x 0.08206Latm/moleK x

295K]/0.971atm = 11.4 liters of B2H6

Answer 2

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